Show that there can't exists $P,Q \in M^{n \times n}(A)$ (ring of matrices over commutative ring $A$) such that $PQ - QP = I$ (the identity).
I thought my solution came too easily, so I wanted a little help to check to see.
Suppose that $PQ - QP = I$, then $PQ = I + QP$, and $Tr(PQ)=Tr(I+QP)$. Now, let's let $p_{ij}$ and $q_{ij}$ but the element $i,j$ of $P$ and $Q$ respectively.
$$Tr(PQ)=\sum_{j=1}^n \sum_{k=1}^n p_{jk}q_{kj} = \sum_{j=1}^n \sum_{k=1}^n q_{kj}p_{jk} = Tr(QP)$$ $$Tr(I+QP)=\sum_{j=1}^n \left ( \left ( \sum_{k=1}^n p_{jk}q_{kj} \right ) + 1 \right ) = \sum_{j=1}^n \sum_{k=1}^n p_{jk}q_{kj} + \sum_{j=1}^n1$$
Where all the sums happen in the ring $A$. This is just the linear property of the trace. So by the properties of the ring operations, we can see that $$Tr(PQ) = Tr(I+PQ) \iff Tr(PQ) - Tr(QP) = Tr(I) \iff 0 = Tr(I)$$
To me, this seems like the contradiction. However, I am just thinking that if the communicative ring $A$ is $\mathbb{Z}/n\mathbb{Z}$, then $\sum^n1=0$. So while the property that $0=Tr(I)$ is not true in general, there does seem to be this special case where it is true.
There can be solutions to this where the characteristic divides $n$.
For instance if $$P=\pmatrix{0&0&0&\cdots&0&0\\1&0&0&\cdots&0&0\\0&2&0&\cdots&0&0 \\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&n-1&0}$$ and $$Q=\pmatrix{0&1&0&\cdots&0&0\\0&0&1&\cdots&0&0\\0&0&0&\cdots&0&0 \\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\0&0&0&0&\cdots&1 \\1&0&0&\cdots&0&0}$$ then $QP-PQ=I$ in characteristic $n$.